Solve for $x$ : $ 8|x + 1| - 9 = -4|x + 1| + 10 $
Solution: Add $ {4|x + 1|} $ to both sides: $ \begin{eqnarray} 8|x + 1| - 9 &=& -4|x + 1| + 10 \\ \\ { + 4|x + 1|} && { + 4|x + 1|} \\ \\ 12|x + 1| - 9 &=& 10 \end{eqnarray} $ Add ${9}$ to both sides: $ \begin{eqnarray} 12|x + 1| - 9 &=& 10 \\ \\ { + 9} &=& { + 9} \\ \\ 12|x + 1| &=& 19 \end{eqnarray} $ Divide both sides by ${12}$ $ \dfrac{12|x + 1|} {{12}} = \dfrac{19} {{12}} $ Simplify: $ |x + 1| = \dfrac{19}{12}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x + 1 = -\dfrac{19}{12} $ or $ x + 1 = \dfrac{19}{12} $ Solve for the solution where $x + 1$ is negative: $ x + 1 = -\dfrac{19}{12} $ Subtract ${1}$ from both sides: $ \begin{eqnarray} x + 1 &=& -\dfrac{19}{12} \\ \\ {- 1} && {- 1} \\ \\ x &=& -\dfrac{19}{12} - 1 \end{eqnarray} $ Change the ${ - 1}$ to an equivalent fraction with a denominator of $12$ $ x = - \dfrac{19}{12} {- \dfrac{12}{12}} $ $ x = -\dfrac{31}{12} $ Then calculate the solution where $x + 1$ is positive: $ x + 1 = \dfrac{19}{12} $ Subtract ${1}$ from both sides: $ \begin{eqnarray} x + 1 &=& \dfrac{19}{12} \\ \\ {- 1} && {- 1} \\ \\ x &=& \dfrac{19}{12} - 1 \end{eqnarray} $ Change the ${ - 1}$ to an equivalent fraction with a denominator of $12$ $ x = \dfrac{19}{12} {- \dfrac{12}{12}} $ $ x = \dfrac{7}{12} $ Thus, the correct answer is $x = -\dfrac{31}{12} $ or $x = \dfrac{7}{12} $.